= Solution
The nuclear <gamma ray> can be absorbed by ground-state <samarium-152>, excite the same level and be re-emitted: nuclear <nuclear resonance fluorescence>. Its <energy> in the absorber rest frame must overlap the nuclear absorption line within the relevant natural and thermal broadening. Atomic radiation from refilling the <K shell> does not have the required transition <energy>.
Let the excitation <energy> be $E_0$ and the nuclear <mass> be $M$. A free stationary emitter loses recoil <energy> while a free stationary absorber requires recoil <energy>:
$$
E_{\mathrm{em}}\simeq E_0-\frac{E_0^2}{2Mc^2},\quad E_{\mathrm{abs}}\simeq E_0+\frac{E_0^2}{2Mc^2},\quad E_{\mathrm{abs}}-E_{\mathrm{em}}\simeq\frac{E_0^2}{Mc^2}.
$$
For $E_0\simeq0.96\ \mathrm{MeV}$ the gap is about $6.5\ \mathrm{eV}$. A <photon> emitted along the daughter recoil receives a positive <Doppler shift> of approximately $E_0p_\nu/(Mc)$, whereas one emitted opposite the recoil is redshifted. Emission before the daughter slows down therefore selects <photons> roughly opposite to the <neutrino>. Comparable <neutrino> and <photon energies> nearly compensate the recoil mismatch; finite line widths and thermal motion permit overlap. Exact compensation need not hold for every atom, and this is not an assumption of recoil-free emission.
Magnetized iron supplies polarized <electrons>. Their <Compton scattering> <cross section> depends on the relative <electron spin> and <photon helicity>, so the iron attenuates the two circular polarizations differently. Reverse the magnetization and measure the resonance-selected count asymmetry $A=(N_+-N_-)/(N_++N_-)$. With a calibrated analysing power $a$ and background correction, $A=aP_\gamma$ determines the sign and magnitude of the circular polarization. A <scintillation counter> such as NaI(Tl), coupled to a <photomultiplier>, determines the <gamma ray> <energy> from calibrated pulse height; a <semiconductor detector> could perform the same spectroscopic function.
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